If _1 = | , * 20 c x&b&b &x&b &a&x , | and _2 = | , * 20 c x&b &x… - Vidyadip

MCQ

If ${\Delta _1} = \left| {\,\begin{array}{*{20}{c}}x&b&b\\a&x&b\\a&a&x\end{array}\,} \right|$ and ${\Delta _2} = \left| {\,\begin{array}{*{20}{c}}x&b\\a&x\end{array}\,} \right|$ are the given determinants, then

A
${\Delta _1} = 3{({\Delta _2})^2}$
✓
$\frac{d}{{dx}}({\Delta _1}) = 3{\Delta _2}$
C
$\frac{d}{{dx}}({\Delta _1}) = 2{({\Delta _2})^2}$
D
${\Delta _1} = 3\Delta _2^{3/2}$

✓

Answer

Correct option: B.

$\frac{d}{{dx}}({\Delta _1}) = 3{\Delta _2}$

b
(b) ${\Delta _1} = \left| {\,\begin{array}{*{20}{c}}x&b&b\\a&x&b\\a&a&x\end{array}\,} \right| = {x^3} - 3abx$ ==> $\frac{d}{{dx}}{\Delta _1} = 3\,({x^2} - ab)$

and ${\Delta _2} = \left| {\,\begin{array}{*{20}{c}}x&b\\a&x\end{array}\,} \right| = {x^2} - ab$ ==> $\frac{d}{{dx}}\,({\Delta _1}) = 3\,({x^2} - ab) = 3{\Delta _2}$.

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